3A

Numpyの行列計算チートシート

はじめに

様々なアルゴリズムを実装する際に漸化式は出るけど実装時にうまくいかない。ということは良くあります。また、Python を使って実装する場合、そのほとんどが行列の取り扱いです。（柔軟であるが故にエラーを吐かずに動いてしまう…笑）

そこで、まずは行列計算のチートシートを作成します。（随時加筆修正していきます。）

目次

  1. 基本事項
  2. 内積（行列積）
  3. 外積
In [1]:
import numpy as np

基本事項

In [2]:
N = 3
A = np.arange(N)
B = np.arange(N).reshape(-1,N) # np.expand_dims(A, axis=0), A[None,:]
C = np.arange(N).reshape(N,-1) # np.expand_dims(A, axis=1), A[:,None]
D = np.arange(N*N).reshape(N,N)
In [3]:
print(f"A = {A}")
print(f"B = {B}")
print(f"C = {C}")
print(f"D = {D}")

A = [0 1 2]
B = [[0 1 2]]
C = [[0]
     [1]
     [2]]
D = [[0 1 2]
     [3 4 5]
     [6 7 8]]
In [4]:
print(f"A.shape = {A.shape}")
print(f"B.shape = {B.shape}")
print(f"C.shape = {C.shape}")
print(f"D.shape = {D.shape}")

A.shape = (3,)
B.shape = (1, 3)
C.shape = (3, 1)
D.shape = (3, 3)
In [5]:
print(f"A.ndim = {A.ndim}")
print(f"B.ndim = {B.ndim}")
print(f"C.ndim = {C.ndim}")
print(f"D.ndim = {D.ndim}")

A.ndim = 1
B.ndim = 2
C.ndim = 2
D.ndim = 2
ndim ex.
scalar $(0)$ 1,2,...
vector $1$ A
matrix $\geq2$ B,C,D

内積（行列積）

vector & scalar / matrix & scalar

  • スカラーが絡むと、各要素とスカラーの積となる。
  • shape の変化はない。
  • どちらから掛けてもエラーは起こらない。
In [6]:
print(f"np.dot(A,2) = {np.dot(A,2)}") # vector & scalar
print(f"np.dot(B,2) = {np.dot(B,2)}") # matrix & scalar
print(f"np.dot(C,2) = {np.dot(C,2)}") # matrix & scalar
print(f"np.dot(D,2) = {np.dot(D,2)}") # matrix & scalar

np.dot(A,2) = [0 2 4]
np.dot(B,2) = [[0 2 4]]
np.dot(C,2) = [[0]
               [2]
               [4]]
np.dot(D,2) = [[ 0  2  4]
               [ 6  8 10]
               [12 14 16]]

vector & vector

  • 内積となる。
  • 結果はスカラーとなる。
  • shape が異なると、エラーが起きる。
In [7]:
print(f"np.dot(A,A) = {np.dot(A,A)}") # vector & vector

np.dot(A,A) = 5
In [8]:
np.dot(A, np.arange(N-1)) # shape=(3,) & shape=(2,)

ValueError: shapes (3,) and (2,) not aligned: 3 (dim 0) != 2 (dim 0)

vector & matrix

  • 行列積となる。
  • ベクトルは、位置によって変換のされ方が異なる。
    • 左がベクトルの場合： np.dot((x,) (y,z)) = np.dot((1,x) (y,z))
    • 右がベクトルの場合： np.dot((y,z) (x,)) = np.dot((y,z) (x,1))
  • 行列積であるので、結果の shape は np.dot((x,y) (y,z)) = (x,z) となる。
  • 上で y が異なるとエラーが生じる。
In [9]:
print(np.dot(A,B)) # np.dot((3,)(1,3))
print(np.dot(A,C)) # np.dot((3,)(3,1))
print(np.dot(B,A)) # np.dot((1,3)(3,))
print(np.dot(C,A)) # np.dot((3,1)(3,))

ValueError: shapes (3,) and (1,3) not aligned: 3 (dim 0) != 3 (dim 0)
[5]
[5]
ValueError: shapes (3,1) and (3,) not aligned: 1 (dim 1) != 3 (dim 0)

matrix & matrix

c = np.dot(a,b)

  • a.shape=(a1,a2,...,aN,X)
  • b.shape=(b1,b2,...,X,bM)
  • c.shape=(a1,a2,...,aN,b1,b2,...,bM)
  • c[i1,i2,...,iN,j1,j2,...,jM] == a[i1,i2,...,iM,:].dot(b[j1,j2,...,jM-1,:,jM])
In [10]:
import itertools
def checkNPdot(shape1, shape2):
    """ 計算可能なshapeの関係を列挙する。 """
    a = np.arange(np.prod(shape1))
    b = np.arange(np.prod(shape2))
    
    for comb1 in itertools.permutations(shape1, len(shape1)):
        for comb2 in itertools.permutations(shape2, len(shape2)):
            try:
                print(f"{comb1}.dot({comb2}) = {(a.reshape(comb1).dot(b.reshape(comb2))).shape}")
            except:
                pass

以下で具体的に見ていく！

np.dot(2D, 3D)

c = np.dot(a,b)

  • a.shape=(i,X)
  • b.shape=(j,X,k)
  • c.shape=(i,j,k)
  • c[i,j,k] == a[i,:].dot(b[j,:,k])
In [11]:
a = np.arange(6).reshape(2,3)
b = np.arange(24).reshape(2,3,4)
c = np.dot(a,b)
In [12]:
print(f"a.shape = {a.shape}")
print(f"b.shape = {b.shape}")
print(f"np.dot(a,b).shape = {c.shape}")

a.shape = (2, 3)
b.shape = (2, 3, 4)
np.dot(a,b).shape = (2, 2, 4)
In [13]:
I,J,K = c.shape
In [14]:
np.all([c[i,j,k]==a[i,:].dot(b[j,:,k]) for i in range(I) for j in range(J) for k in range(K)])
Out[14]:
True

np.dot(3D,2D)

c = np.dot(a,b)

  • a.shape=(i,j,X)
  • b.shape=(X,k)
  • c.shape=(i,j,k)
  • c[i,j,k] == a[i,j,:].dot(b[:,k])
In [15]:
a = np.arange(24).reshape(4,2,3)
b = np.arange(6).reshape(3,2)
c = np.dot(a,b)
In [16]:
print(f"a.shape = {a.shape}")
print(f"b.shape = {b.shape}")
print(f"np.dot(a,b).shape = {c.shape}")

a.shape = (4, 2, 3)
b.shape = (3, 2)
np.dot(a,b).shape = (4, 2, 2)
In [17]:
I,J,K = c.shape
In [18]:
np.all([c[i,j,k] == a[i,j,:].dot(b[:,k]) for i in range(I) for j in range(J) for k in range(K)])
Out[18]:
True

np.dot(3D,3D)

c = np.dot(a,b)

  • a.shape=(i,j,X)
  • b.shape=(k,X,l)
  • c.shape=(i,j,k,l)
  • c[i,j,k,l] == a[i,j,:].dot(b[k,:,l])
In [19]:
a = np.arange(24).reshape(2,3,4)
b = np.arange(60).reshape(3,4,5)
c = np.dot(a,b)
In [20]:
print(f"a.shape = {a.shape}")
print(f"b.shape = {b.shape}")
print(f"np.dot(a,b).shape = {c.shape}")

a.shape = (2, 3, 4)
b.shape = (3, 4, 5)
np.dot(a,b).shape = (2, 3, 3, 5)
In [21]:
I,J,K,L = c.shape
In [22]:
np.all([c[i,j,k,l] == a[i,j,:].dot(b[k,:,l]) for i in range(I) for j in range(J) for k in range(K) for l in range(L)])
Out[22]:
True

np.dot(2D,4D)

c = np.dot(a,b)

  • a.shape=(i,X)
  • b.shape=(j,k,X,l)
  • c.shape=(i,j,k,l)
  • c[i,j,k,l] == a[i,:].dot(b[j,k,:,l])
In [23]:
a = np.arange(6).reshape(2,3)
b = np.arange(120).reshape(2,4,3,5)
c = np.dot(a,b)
In [24]:
print(f"a.shape = {a.shape}")
print(f"b.shape = {b.shape}")
print(f"np.dot(a,b).shape = {c.shape}") # c[i,j,k,l] == a[i,j,:].dot(b[k,:,l])

a.shape = (2, 3)
b.shape = (2, 4, 3, 5)
np.dot(a,b).shape = (2, 2, 4, 5)
In [25]:
I,J,K,L = c.shape
In [26]:
np.all([c[i,j,k,l] == a[i,:].dot(b[j,k,:,l]) for i in range(I) for j in range(J) for k in range(K) for l in range(L)])
Out[26]:
True

np.dot(3D,4D)

c = np.dot(a,b)

  • a.shape=(i,j,X)
  • b.shape=(k,l,X,m)
  • c.shape=(i,j,k,l,m)
  • c[i,j,k,l,m] == a[i,j,:].dot(b[k,l,:,m])
In [27]:
a = np.arange(24).reshape(2,3,4)
b = np.arange(360).reshape(3,5,4,6)
c = np.dot(a,b)
In [28]:
print(f"a.shape = {a.shape}")
print(f"b.shape = {b.shape}")
print(f"np.dot(a,b).shape = {c.shape}")

a.shape = (2, 3, 4)
b.shape = (3, 5, 4, 6)
np.dot(a,b).shape = (2, 3, 3, 5, 6)
In [29]:
I,J,K,L,M = c.shape
In [30]:
np.all([c[i,j,k,l,m] == a[i,j,:].dot(b[k,l,:,m]) for i in range(I) for j in range(J) for k in range(K) for l in range(L) for m in range(M)])
Out[30]:
True

np.dot(4D,4D)

c = np.dot(a,b)

  • a.shape=(i,j,k,X)
  • b.shape=(l,m,X,n)
  • c.shape=(i,j,k,l,m,n)
  • c[i,j,k,l,m,n] == a[i,j,k,:].dot(b[l,m,:,n])
In [31]:
a = np.arange(120).reshape(2,3,4,5)
b = np.arange(360).reshape(3,4,5,6)
c = np.dot(a,b)
In [32]:
print(f"a.shape = {a.shape}")
print(f"b.shape = {b.shape}")
print(f"np.dot(a,b).shape = {c.shape}")

a.shape = (2, 3, 4, 5)
b.shape = (3, 4, 5, 6)
np.dot(a,b).shape = (2, 3, 4, 3, 4, 6)
In [33]:
I,J,K,L,M,N = c.shape
In [34]:
np.all([c[i,j,k,l,m,n] == a[i,j,k,:].dot(b[l,m,:,n]) for i in range(I) for j in range(J) for k in range(K) for l in range(L) for m in range(M) for n in range(N)])
Out[34]:
True

外積

  • c = np.outer(a,b)
  • c[i][j] = a.reshape(-1,)[i]*b.reshape(-1,)[j]
In [35]:
c = np.outer(a,b)
In [36]:
print(f"a.shape = {a.shape}")
print(f"b.shape = {b.shape}")

a.shape = (a1,a2,...,aN)
b.shape = (b1,b2,...,bM)
In [37]:
Na = np.prod(a.shape)
Nb = np.prod(b.shape)
In [38]:
print(f"Na = {Na}")
print(f"Nb = {Nb}")

Na = a1*a2*・・・*aN
Nb = b1*b2*・・・*bM
In [39]:
print(c.shape)

(Na, Nb)
In [40]:
np.all([c[i][j] == a.reshape(-1,)[i]*b.reshape(-1,)[j] for i in range(Na) for j in range(Nb)])
Out[40]:
True
In [ ]:
hidden
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